Question

Let the curves $\frac{x^2}{a^2}+\frac{y^2}{b^2}=4$ and $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ have same foci. If the eccentricity of the given curves are $e_1$ and $e_2$ respectively, then

• A. $e_2=\sqrt{\frac{32}{17}}$
• B. $e_1=\sqrt{\frac{2}{17}}$
• C. $e_2=\sqrt{\frac{8}{5}}$
• D. $e_1=\sqrt{\frac{2}{5}}$

Answer 1

Answer: (C), (D)

The correct options are
C $e_2=\sqrt{\frac{8}{5}}$
D $e_1=\sqrt{\frac{2}{5}}$

From the equation of the hypberbola, the foci of the hyperbola lie on $x-$ axis
$\therefore$ Ellipse will be horizontal ellipse.
Here $e_1$ is the eccentricity of ellipse and $e_2$ is eccentricity of hyperbola.
Now, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=4$
$\Rightarrow \frac{x^2}{(2a{)}^2}+\frac{y^2}{(2b{)}^2}=1$
$\Rightarrow e_1^2=1-\frac{(2b{)}^2}{(2a{)}^2}$
$\Rightarrow e_2^2=\frac{b^2}{a^2}+1$
$\Rightarrow e_1^2+e_2^2=2\cdots \left(1\right)$
As foci of the ellipse will be same as that of hyperbola
$\therefore 2a{e}_1=a{e}_2\Rightarrow e_2=2e_1\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we get
$e_1=\sqrt{\frac{2}{5}},e_2=\sqrt{\frac{8}{5}}$